IPMAT IndoreAlgebra > Easy{1}\{1\}{1}{1,−1}\{1,-1\}{1,−1}{1,0}\{1,0\}{1,0}{1,0,−1}\{1,0,-1\}{1,0,−1}✅ Correct Option: 2Related questions:IPMAT Indore 2021Suppose that a real-valued function f(x)f(x)f(x) of real numbers satisfies f(x+xy)=f(x)+f(xyf(x + xy) = f(x) + f(xyf(x+xy)=f(x)+f(xy) for all real x,y,x, y,x,y, and that f(2020)=1f(2020) = 1f(2020)=1. Compute f(2021)f(2021)f(2021).IPMAT Indore 2019A real-valued function fff satisfies the relation f(x)f(y)=f(2xy+3)+3f(x+y)−3f(y)+6yf(x)f(y) = f(2xy + 3) + 3f(x + y) - 3f(y) + 6yf(x)f(y)=f(2xy+3)+3f(x+y)−3f(y)+6y, for all real numbers xxx and yyy, then the value of f(8)f(8)f(8) isIPMAT Indore 2020Given f(x)=x2+log3xf(x) = x^2 + \log_3 xf(x)=x2+log3x and g(y)=2y+f(y)g(y) = 2y + f(y)g(y)=2y+f(y), then the value of g(3)g(3)g(3) equals